Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 300005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
    	if(c == '-') f = -1;
    	c = getchar();
    }
    while(c >= '0' && c <= '9') {
    	res = res * 10 +c - '0';
    	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    	out(x / 10);
    }
    putchar('0' + x % 10);
}
int N;
int p[MAXN],pos[MAXN];
int pre[MAXN],f[MAXN],ans[MAXN];
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) {
        read(p[i]);pos[p[i]] = i;
    }
    for(int i = N - 1 ; i >= 1 ; --i) {
        if(ans[i + 1] == 0) {
            if(pos[i] < pos[i + 1]) ans[i] = 0;
            else {ans[i] = 1;f[0] = i + 1;}
        }
        else {
            int t = ans[i + 1] - 1;
            int a[3] = {i,i + 1,f[t]};
            sort(a,a + 3,[](int c,int d){return pos[c] < pos[d];});
            while(a[0] != f[t]) {
                swap(a[2],a[0]);swap(a[1],a[2]);
            }
            if(a[1] == i) {
                ans[i] = ans[i + 1];
            }
            else {
                ans[i] = ans[i + 1] + 1;
                f[t + 1] =  i + 1;
            }
        }
    }
    out(ans[1]);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}